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\(B=\dfrac{6^{30}\cdot3^{11}\cdot7-2^{31}\cdot5\cdot3^{40}}{2^{28}\cdot17\cdot3^{38}-6^{27}\cdot3^{11}}\)
\(=\dfrac{2^{30}\cdot3^{41}\cdot7-2^{31}\cdot5\cdot3^{40}}{2^{28}\cdot17\cdot3^{38}-2^{27}\cdot3^{38}}\)
\(=\dfrac{2^{30}\cdot3^{40}\left(3\cdot7-2\cdot5\right)}{2^{27}\cdot3^{38}\left(2\cdot17-1\right)}\)
\(=\dfrac{2^3\cdot3^2\cdot\left(21-10\right)}{34-1}=\dfrac{2^3\cdot3^2\cdot11}{33}=\dfrac{2^3\cdot3^2}{3}=2^3\cdot3=24\)
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a: \(\dfrac{-3}{25}\cdot\dfrac{4}{17}\cdot\dfrac{5}{-3}\cdot\dfrac{17}{2}=\dfrac{4}{17}\cdot\dfrac{17}{2}\cdot\dfrac{3}{25}\cdot\dfrac{5}{3}=2\cdot\dfrac{1}{5}=\dfrac{2}{5}\)
b: \(=-\dfrac{8}{15}\cdot\dfrac{13}{16}\cdot\dfrac{30}{13}=\dfrac{-8}{15}\cdot\dfrac{15}{8}=-1\)
c: \(=\dfrac{3}{11}\left(\dfrac{2}{9}\cdot\dfrac{-8}{33}+\dfrac{7}{9}\right)=\dfrac{3}{11}\cdot\dfrac{215}{297}=\dfrac{2015}{1089}\)
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1. Tìm x, biết:
a) \(\frac{4}{x}=\frac{8}{6}\). Ta có: \(\frac{8}{6}=\frac{8:2}{6:2}=\frac{4}{3}\Rightarrow x=3\)
b) \(\frac{3}{x-5}=\frac{4}{x+2}\). Ta có: \(5-2=3\)
\(\Rightarrow x=\left(3.5\right)+\left(4.2\right)+3=15+8+3=26\)
c) \(\frac{x}{-2}=\frac{-8}{x}\Rightarrow x=\left(-8\right):\left(-2\right)=4\)
2. Rút gọn
a) \(\frac{2^4.5^2.11^2.7}{2^3.5^3.7^2.11}\Leftrightarrow\frac{2^3.2^1.5^2.11.11.7}{2^3.5^2.5^1.7.7.11}\Leftrightarrow\frac{2^1.11}{5^1.7}=\frac{22}{35}\)
b) Tương tự
c) Tương tự
\(\frac{4}{x}=\frac{8}{6}\Rightarrow x=\frac{4.6}{8}=3\)
\(\frac{3}{x-5}=\frac{-4}{x+2}\Rightarrow3\left(x+2\right)=-4\left(x-5\right)\)
\(\Rightarrow3x+6=-4x+20\)
\(\Rightarrow3x+4x=20-6\)
\(\Rightarrow7x=14\)
\(\Rightarrow x=2\)
\(\frac{x}{-2}=\frac{-8}{x}\Rightarrow x^2=\left(-8\right)\left(-2\right)=16\Rightarrow x=\pm4\)